Share & grow the world's knowledge!
We want to connect the people who have knowledge to the people who need it, to bring together people with different perspectives so they can understand each other better, and to empower everyone to share their knowledge.
A straight conducting rod of length l and mass is suspended in a horizontal plane by a pair of flexible strings in a magnetic field of magnitude B. To remove the tension in the supporting strings, the magnitude of current in the wire is :
To remove tension in the strings, the magnetic force must balance the gravitational force: Bil = mg Rearranging for current I: I = mg/BL Thus, the magnitude of current needed to make the net force zero is: I = mg/BL Answer: (C) mg/IB For more visit here: https://www.tiwariacademy.com/ncert-solutionsRead more
To remove tension in the strings, the magnetic force must balance the gravitational force:
Bil = mg
Rearranging for current I:
I = mg/BL
Thus, the magnitude of current needed to make the net force zero is:
I = mg/BL
Answer: (C) mg/IB
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
The current sensitivity of a galvanometer increases by 20%. If its resistance also increases by 25%, the voltage sensitivity will
The voltage sensitivity of a galvanometer is given by: Voltage Sensitivity = Current Sensitivity/R If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is: 1.2/1.25 = 0.96 This indicates a 4% decrease in voltage sensitivity. Answer: (D) DecreRead more
The voltage sensitivity of a galvanometer is given by:
Voltage Sensitivity = Current Sensitivity/R
If current sensitivity increases by 20% and resistance increases by 25%, the net change in voltage sensitivity is:
1.2/1.25 = 0.96
This indicates a 4% decrease in voltage sensitivity.
Answer: (D) Decrease by 4%.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
A current of 5 A is flowing from east to west In a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due south on the table is :
The magnetic field around a long, straight current-carrying wire is given by: B = μ0I/2πr Substituting values: B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is: (B) 1 × 10⁻⁵ T acting upwarRead more
The magnetic field around a long, straight current-carrying wire is given by:
B = μ0I/2πr
Substituting values:
B = 4π × 10⁻⁷ × 5/2π × 0.10 = 1 × 10⁻⁵ T
Using the right-hand thumb rule, the magnetic field at a point due south acts upward. Therefore, the correct answer is:
(B) 1 × 10⁻⁵ T acting upwards.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
There are unifonn electric and magnetic fields in a region pointing along X-axis. An α-particle is projected along Y-axis with a velocity v. The shape of the trajectory will be :
An α-particle projected along the Y-axis with velocity v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along theRead more
An α-particle projected along the Y-axis with velocity
v in the presence of uniform electric and magnetic fields along the X-axis experiences a magnetic force perpendicular to both velocity and magnetic field, causing circular motion in the YZ plane. The electric field imparts acceleration along the X-axis, resulting in a helical trajectory with its axis parallel to the X-axis.
Answer: (C) Helical with its axis parallel to X-axis.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/
A current 10 A is flowing from east to west in a long straight wire kept on a horizontal table. The magnetic field developed at a distance of 10 cm due north on the table is
The magnetic field around a long, straight current-carrying wire is given by Ampere’s circuital law as: B = μ0I/2πr .Here: μ0 4π × 10⁻⁷Tm/A (permeability of free space) I =10 A (current) r = 0.10m (distance) Substituting values: B = 4π × 10⁻⁷× 10/2π × 0.10 = 2 × 10⁻⁵T To determine the direction ofRead more
The magnetic field around a long, straight current-carrying wire is given by Ampere’s circuital law as:
B = μ0I/2πr
.Here:
μ0 4π × 10⁻⁷Tm/A (permeability of free space)
I =10 A (current) r = 0.10m (distance)
Substituting values:
B = 4π × 10⁻⁷× 10/2π × 0.10 = 2 × 10⁻⁵T
To determine the direction of the magnetic field, use the right-hand thumb rule: Point the thumb in the direction of current (east to west) and curl the fingers around the wire. At a point due north, the magnetic field points downward. Hence, the correct answer is:
(A) 2 × 10⁻⁵ T, acting downwards.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-4/